Each of two large conducting parallel plates | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When a charge $Q$ is given to one of two large parallel conducting plates,the charge redistributes itself on both sides of each plate such that th

Vedclass · Accepted Answer

$\frac{Q}{2 A \varepsilon_0}$

Vedclass · Answer

(B) When a charge $Q$ is given to one of two large parallel conducting plates,the charge redistributes itself on both sides of each plate such that the electric field inside the conducting material is zero.Let the charges on the four surfaces (from left to right) be $q_1, q_2, q_3,$ and $q_4$.For the first plate: $q_1 + q_2 = Q$ and for the second plate: $q_3 + q_4 = 0$.Since the plates are large,the electric field due to a surface with charge density $\sigma$ is $E = \frac{\sigma}{2 \varepsilon_0} = \frac{q}{2 A \varepsilon_0}$.Inside the conducting material of the plates,the net electric field must be zero.Applying this condition,we find that the charges on the inner surfaces are $q_2 = \frac{Q}{2}$ and $q_3 = -\frac{Q}{2}$,while the outer surfaces have $q_1 = \frac{Q}{2}$ and $q_4 = \frac{Q}{2}$.The electric field at a point between the plates is due to the inner surface charges $q_2$ and $q_3$.$E_{	ext{net}} = E_2 + E_3 = \frac{q_2}{2 A \varepsilon_0} + \frac{|q_3|}{2 A \varepsilon_0} = \frac{Q/2}{2 A \varepsilon_0} + \frac{Q/2}{2 A \varepsilon_0} = \frac{Q}{4 A \varepsilon_0} + \frac{Q}{4 A \varepsilon_0} = \frac{Q}{2 A \varepsilon_0}$.

Each of two large conducting parallel plates has one-sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral,then the electric field at a point in between the plates is given by

Similar Questions

Three identical metal plates with large surface areas are kept parallel to each other as shown in the figure. The leftmost plate is given a charge $Q$,the rightmost a charge $-2Q$,and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate.

$A$ circle of radius $a$ has a linear charge density given by $\lambda = \lambda_0 \cos^2 \theta$ on its circumference. What will be the total charge on the circle?

$A$ uniformly charged conducting sphere of $2.4 \, m$ diameter has a surface charge density of $80.0 \, \mu C m^{-2}$. The charge on the sphere is nearly

The surface charge density $\sigma$ on a charged conducting sphere of radius $R$ in terms of the electric field intensity $E$ at a distance $r$ $(r > R)$ in free space is (where $\varepsilon_{0}$ is the permittivity of free space):

Vedclass Test Series

Exam Paper Generator

Online Exam Module